Introduction to FEM

INTRODUCTION TO FINITE ELEMENT METHOD (FEM)

1. INTRODUCTION

Preliminary design: concept

design

ENGINEERING DEPARTMENT: To develop new products and/or manufacturing processes

Product calculation: - Component dimensioning

- Design verification - Material selection

Process calculation: - Process parameters

- Tooling design

INDUSTRIALISATION: - Development of manufacturing

drawings - Manufacturing process

definition - Development of manufacturing

tools - Production

2

1. INTRODUCTION: CALCULATION METHODS

- Analytic method:

- Consists on the use of analytic equation to represent the behaviour of a physical problem (As exercises solved by hand in previous chapters or other subjects, heat transfer, material resistance, dynamics, vibrations,…)

- Advantages: relatively fast to be solved

- Limitations: Hard to represent complex phenomena in real components, not always applicable.

3

1. INTRODUCTION: CALCULATION METHODS

- Numeric methods (Finite Element Methods FEM)

- To divide a complex problem into many simple problems (elements)

- Problem solution by numeric methods (Newton-Raphson) by using iterations and increments.

- Advantages: Capability to solve complex problems.

- Limitations: Time expensive resolution method, the use of computers is required.

4

1. INTRODUCTION: FORMULATION TYPES

- Implicit:

- Explicit:

- Tries to obtain the structural equilibrium for each time increment. - More sophisticate algorithms higher time increments (FASTER). - High precision - Convergence problems when solving non-linear phenomena: hard variations in boundary condition, material behaviour, loads, contacts,…

- Does not need iterations, just time increments (Does not try to get the exact solution) - No convergence problems - Utilizes constant time increments - High calculation time - Recommendable to solve non-linear problems.

5

1. INTRODUCTION: APLICATIONS

Solid mechanics:

- Structural linear calculations (linear static, linear dynamics) IMPLICIT - Plasticity range calculation (no-linear quasi-static or dynamics) EXPLICIT

Fluid mechanics:

- Linear calculation (wind tunnel example) IMPLICIT

- Non-linear calculations (atmospheric phenomena, turbulence, wind,...) EXPLICIT

Thermodynamics: (linear problems-IMPLICIT)

Multiphysics: thermo mechanic, thermo fluidic, fluid structure interaction…

6


Solid mechanics: Structural static calculation

Set-up

Aluminium sheet bulge-test

Design

vs.

FEM

IMPLICIT

7


8

Solid mechanics: eigenvalues

Trunk door

First mode17Hz

Solid mechanics: Forming processes

Punching

IMPLIT

EXPLICIT


9

Solid mechanics: Forming processes


10

Vc=300 m.min-1 Vc≥600 m.min-1

Solid mechanics: Machining process

EXPLICIT

11

Fluid mechanics: linear and non-linear examples

Air flow simulation F1

Hurricane simulation

EXPLICIT IMPLICIT


12

Thermodynamics:

Turbine heat transfer simulation Tube and die temperature

pattern simulation IMPLICIT

IMPLICIT


2. FEM THEORY: STATIC IMPLICIT LINEAR CALCULATION

13

- Definition of a finite element

Geometrical definition (element shape): - Composed by nodes Physic definition (element type): - Degrees of freedom (DOF) - Analytic formulation of the element (mechanical field resolution, thermal fields,…)

u v w θx θy θz

6 Degrees of Freedom

(DOF)

x y

z

u v

w

θx

θy

θz

u: Linear displacement in X v: Linear displacement in y w: Linear displacement in z θx: Rotation with respect X θy : Rotation with respect Y θz : Rotation with respect Z

14

Interpolation order: linear (1st order) y quadratic (2nd order)

- Element types:

v1 v2 v1

v2 v3

Linear interpolation Quadratic interpolation

V(x)=mx+b V(x)=ax2+bx+c

x x v1 θ1 v2 θ2

δ= Nodal displacement vector of a first order beam element in 2D de

*


15

- Mechanical field calculation: Motion differential equation

[M]{δ} + [C]{δ}+[k]{δ} ={Fext} . .. [M]: Mass matrix

[C]: Damping matrix [k]: Stiffness matrix {δ}: Displacement vector

{δ}: Velocity vector .

{δ}: Acceleration vector ..

{Fext}: External load vector

- Mechanical field: STATIC

Acceleration = 0 Velocity = 0

[M]{δ} + [C]{δ}+[k]{δ} ={Fext} . ..

[k]{δ} ={Fext}

5 unknowns and 5 equations


16

- Stress field calculation through nodal displacement.

.

. ] N,....,N,N [=}{

n

1

n21e

δ

δ

δ

}{] [ = }{

w

v

u

.

x 0

z

y

z

0

0 x

y

z

0 0

0 y

0

0 0 x

=

zx

yz

xy

z

y

x

δε

γ

γ

γ

ε

ε

ε

∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

INTERPOLATION displacement at any point of the structure

Strain through local displacement

( ) ( )

( )( )

( )( )

( )

( )

−

−

−−

−−

+⋅⋅−=

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

γγγεεε

ν

ν

νννν

νννννν

ννE

τττσσσ

22100000

02210000

00221000

000100010001

121

Generalised Hooke’s law Stress calculation through local strain ε1

ε3

ε2


17

- INTERPOLATION FUNCTIONS: Determination of the displacement at any point of the structure.


{ } [ ]{ }e*

e δNδ =

{ } [ ]

=

1

,...,, 21

δ

δδ

nne NNN

So, the displacement at any point of a determined element is obtained:

Where [Nk] represents the contribution of node k’s displacement in the total displacement of any determined point.

{ } [ ]{ } [ ]{ } [ ]{ }nnNNN δδδδ +++= ...2211

= displacement vector at any point in a determined element

= nodal displacement vector of a determined element

= interpolation function matrix

{ }eδ

{ }e*δ

[ ]N

= interpolation function of the nodal displacement a determined node i.

= displacement vector at a determined node i.

{ }iδ

[ ]iN

18

- STIFFNESS MATRIX:


In order to cause the nodal displacements and consequently deform the element it is necessary the presence of nodal forces

Definition: The stiffness coefficient Kij represents the necessary force to apply to a certain degree of freedom i to obtain an unitary displacement of the degree of freedom j being 0 the influence in the displacement of the rest of the degrees of freedom

{ }*δ{ }*f

j

n

jiji Kf δ⋅=∑

=1n= number of DOF

nnnnnn

nn

fKKK

fKKK

=⋅++⋅+⋅

=⋅++⋅+⋅

δδδ

δδδ

...

...

2211

11212111

Writing in matrix form: { } [ ] { }eee Kf ** δ=

[K]e = Stiffness matrix of the element

19


DETERMINATION OF THE STIFFNESS MATRIX OF A FINITE ELEMENT

Relation between nodal forces an nodal displacements:

Based on CAPLEYRON theory, the external work of the nodal forces is represented:

The internal deformation energy caused by the nodal displacements:

{ } [ ]{ }** δKf =

{ } { }**

21 fw Tδ=

{ } { } dvu T ⋅= ∫ σε21

{ } { } [ ]{ } [ ]{ }{ } [ ]{ } [ ][ ]{ }*

**

δεσ

δδδε

BDD

BN

==

=∂=∂=As: { } [ ]{ }

[ ][ ]{ }{ }

dvBDBuTTv

TT⋅= ∫

σε

δδ **

21

Being uw =

{ } [ ]{ }**

21 δδ Kw T

=

{ } [ ]{ } { } [ ] [ ][ ] { }****

21

21 δδδδ

⋅= ∫ dvBDBK

v

TTT

[ ] [ ] [ ][ ] dvBDBKv

T ⋅= ∫ STIFFNESS MATRIX

20


- DETERMINATION OF THE STIFFNESS MATRIX IN GLOBAL COORDINATES

Transformation matrix

[ ]

=

zyx

zyx

zyx

cccbbbaaa

T

From local coordinate system of the element

To global coordinate system

{ } [ ]{ }** δδ T= { } [ ]{ }** δKf = { } [ ]{ }** δKf =

{ } [ ] { } [ ] [ ]{ } [ ] [ ][ ]{ }**** δδ TKTKTfTf TTT ===

[K] in GLOBAL coord. system

{ } [ ]{ }** fTf =

{ } [ ] { }** fTf T= [ ] [ ] [ ][ ]TKTK T=

21

- INTERPOLATION FUNCTION OF A TRUSS ELEMENT


Truss element:

- 2 node one-dimensional element (2DOF)

- Only allows to calculate tractive-compressive condition

Determination of the interpolation function 21 ,

,,,

uuji

zyxNodal displacement vector { } { }21,uuT =δ

2 G.D.L 1st order equation xaaxu .)( 10 +=

x

y

1 2

2u1u

L i j

)(xu

Local axis Element nodes Nodal displacements

1)0( uu =

2)( ulu = } laauau

102

01

+== }

=

1

0

2

1

101

aa

luu

211

10

11 ul

ul

a

ua

+−=

= }

−

=

2

1

1

0

1101

uul

laa [ ]

−=

−=

2

1

2

1 ,11101

,1uu

lx

lx

uu

llxu

[ ]

=2

121, u

uNNu

lxN −=11 l

xN =2

22

- STIFFNESS MATRIX OF A TRUSS ELEMENT


Truss element: 2DOF

x

y

1 2

2u1u

L

{ }yx

yx uu

,2

1,

*

=δ

Local coordinate system

[ ] [ ] [ ][ ]∫=v

T dvBDBk .

[ ]

−=

=2

1

2

121, 1

uu

lx

lx

uu

NNu yx

[ ] [ ]21 ,NNN =

lxN

lxN

=

−=

2

1 1

Stiffness matrix obtaining formula:

[ ][ ] [ ]

−=

∂∂

∂∂

=

∂∂

=∂∂

=2

1

2

121

2

121

1,1,,uu

lluu

xN

xN

uu

NNxx

u

BB

x

ε

[ ] ∫∫

−

−=

−

−=

−

−=

l

v lSEdx

ll

llESdvll

E

l

lK0

22

22

1111..

11

11

..1,11

1

23

- STIFFNESS MATRIX OF A TRUSS ELEMENT


[ ]

−

−=

1111.

eLSEk

Stiffness matrix of TRUSS element in local coordinate system

Displacement vector in global axis: y

x

x

y

1u

ϑ

- RIGIDITY MATRIX OF A TRUSS ELEMENT IN GLOBAL COORD. SYSTEM

yxYX

v

u

vuvu

,

1

1

,2

2

1

1

0

0

cossin00sincos0000cossin00sincos

−

−

=

θθθθ

θθθθ

24


Naming

==

ϑλϑµ

cossen

- STIFFNESS MATRIX OF A TRUSS ELEMENT IN GLOBAL COORD. SYSTEM

[ ] [ ] [ ] [ ]

[ ]

−

−

−

−

−

−

=

=

λµµλ

λµµλ

λµµλ

λµµλ

0000

0000

0000010100000101

0000

0000

.lSEK

TKTK

e

Te

[ ]

−−

−−

−−

−−

=

22

22

22

22

.

µµλµλµ

λµλλµλ

µλµµµλ

λµλλµλ

lSEK e

Stiffness matrix of a TRUSS element in GLOBAL coordinate system

25

- INTERPOLATION FUNCTION OF A BEAM ELEMENT


2v1v

z

x

y

1ϑ 2ϑ

Beam element:

- 2 node unidimensional element (4DOF)

- Only allows to calculate behind loading condition

2

2

1

1

ϑ

ϑv

v

{ }=Tδ

=

2

2

1

1

4321

4321

ϑ

ϑϑ v

v

dxdN

dxdN

dxdN

dxdN

NNNNv

+−−+−+−

+−

−

+−

+

−

=

2

2

1

1

2

2

3

2

22

22

2

2

3232

2

3232

32,66,341,66

,23,2,231

ϑ

ϑϑ v

v

lx

lx

lx

lx

lx

lx

lx

lx

lx

lx

lx

lx

lx

lxx

lx

lx

v

Beam element interpolation function:

26

- INTERPOLATION FUNCTION OF A COMPLETE BEAM ELEMENT


2v1v

z

x

y

1ϑ 2ϑ

Complete Beam element:

- 2 node one-dimensional element (6DOF)

- Only allows to calculate behind loading condition

2

2

2

1

1

1

ϑ

ϑ

vu

vu

{ }=Tδ

=

2

2

2

1

1

1

6543

6543

21

00

000000

ϑ

ϑ

ϑvu

vu

dxdN

dxdN

dxdN

dxdN

NNNNNN

vu

+−−+−+−

+−

−

+−

+

−

−

=

2

2

2

1

1

1

2

2

3

2

23

22

2

2

3232

2

3232

32660341660

23022310

00001

ϑ

ϑ

ϑvu

vu

lx

lx

lx

lx

lx

lx

lx

lx

lx

lx

lx

lx

lx

lxx

lx

lx

lx

lx

vu

Complete Beam element interpolation function:

2u1u

27

- STIFFNESS MATRIX OF A BEAM ELEMENT


BEAM element: 4 DOF

{ }

yx

yx v

v

,2

2

1

1

,*

=

θ

θδ

Local coordinate system

[ ] [ ]4321 ,,, NNNNN =

Beam deflection 2

2

dd

dd

xvy

xyx −=−=

θε

[ ]{ }( ) [ ]{ }**2

2

2

2

dd

dd δδε BN

xy

xvyx =−=−=

[ ] [ ] [ ][ ] == ∫ vBDBkv

T d

∫∫

+−−+−+−

+−

−

+−

+−

=s

l

syxlx

llx

llx

llx

l

lx

l

lx

l

lx

l

lx

l

E dd62,126,64,126

62

126

64

126

2232232

0

2

32

2

32

xv

ddbeing =θ

[ ]

+−−+−+−−= 232232 62,126,64,126

lx

llx

llx

llx

lyB

28

- STIFFNESS MATRIX OF A BEAM ELEMENT


Stiffness matrix of BEAM element in local coordinate system

Displacement vector in global axis: y

x

x

y

1u

ϑ

- STIFFNESS MATRIX OF A BEAM ELEMENT IN GLOBAL COORD. SYSTEM

yxYX v

uv

u

vuvu

,2

2

1

1

,2

2

1

1

cossin00sincos0000cossin00sincos

−

−

=

θθθθ

θθθθ

29


Naming

==

ϑλϑµ

cossen

- STIFFNESS MATRIX OF A BEAM ELEMENT IN GLOBAL COORD. SYSTEM

Stiffness matrix of a BEAM element in GLOBAL coordinate system

[ ]

−

−

−

−−−

−

−

−

−

=

1000000000000000010000000000

460260

61206120

000000

260460

61206120

000000

1000000000000000010000000000

.

22

2323

22

2323

λµµλ

λµµλ

λµµλ

λµµλ

llll

llll

llll

llll

IEk ze

[ ]

−−−

−−−

−

−

−

=

llllll

lllll

llll

lll

ll

l

IEk ze

466266

121261212

1261212

466

1212

12

.

223

2332

233

2323

23

22

233

23

µµµ

λλµλλλµ

µµλµµ

λµ

λλ

µ

30


Naming

==

ϑλϑµ

cossen

- STIFFNESS MATRIX OF A COMPLETE BEAM ELEMENT IN GLOBAL COORD. SYSTEM

2 23

2 23 3

2 2

2 2 2 23 3 2 3

2 2 2 23 3 2 3 3

2 2

12

12 12 .

6 6 4

12 12 6 12

12 12 6 12 12

6 6 2

Te e

EI EAL L

EI EA EI EA syL L L L

EI EI EIL L L

EI EA EI EA EI EI EAL L L L L L L

EI EA EI EA EI EI EA EI EAL L L L L L L L L

EI EIL L

µ λ

µλ µλ λ µ

µ λ

µ λ µλ µλ µ µ λ

µλ µλ λ µ λ µλ µλ λ µ

µ λ

+

− + +

−

= =

− − − +

− − − − − + +

−

K T k T

2 2

6 6 4

cossin

EI EI EI EIL L L L

µ λ

λ θµ θ

−

==

31

2. FEM THEORY: STATIC IMPLICIT LINEAR CALCULATION - DETERMINATION THE EQUIVALENT NODAL LOAD VECTOR

In a real problem different type of external loads can be found:

- Punctual forces

- Moments

- Distributed loads

f*

f

=

For FEM modelling all external load should be applied in the element nodes

- Punctual forces

- Moments

- Distributed loads

NECESITY TO OBTAIN AN EQUIVALENT SYSTEM BASED IN NODAL LOADS

32

2. FEM THEORY: STATIC IMPLICIT LINEAR CALCULATION - DETERMINATION THE EQUIVALENT NODAL LOAD VECTOR

{ } { }∫ ⋅=s

T dsfw δ21

1

The external work due to all the external load applied to the system is given by

By using the interpolation functions:

Thus the work of the equivalent system can be written as:

{ } [ ] { } { } [ ] { }∫∫ ⋅=⋅=s

TT

s

TT dsfNdsfNw **1 2

121 δδ

{ } { }**2 2

1 fw Tδ=

21 ww = { } [ ] { } { } { }***

21

21 fdsfN T

s

TTδδ =⋅∫

{ } [ ] { }∫ ⋅=s

T dsfNf *

33

2. FEM THEORY: STATIC IMPLICIT LINEAR CALCULATION DETERMINATION OF THE STRESS/STRAIN CONDITION:

Once, the nodal displacement vector of the studied system is solved the stress/strain condition at any point can be obtained.

STEP 1: STRAIN DETERMINATION AT A CERTAIN POINT

{ } [ ]{ }δε ∂=

.

. ] N,....,N,N [=}{

n

1

n21e

δ

δ

δ

∂∂

∂∂

∂∂

∂∂∂

∂∂

∂∂

∂∂

∂∂

∂

=

wvu

xz

yz

xy

z

y

x

zx

yz

xy

z

y

x

0

0

0

00

00

00

γγγεεε

Determination of the elongation at the selected point Strain vector determination

{ } [ ]{ } [ ]{ }** δδε BN =∂=

34


STEP 2: STRESS DETERMINATION AT A CERTAIN POINT The relation between the strain and the stress in the linear elastic domain is given by the generalised Hooke’s law:

( )[ ]

( )[ ]

( )[ ]yxzz

xzyy

zyxx

E

E

E

σσυσε

σσυσε

σσυσε

+−=

+−=

+−=

1

1

1 ( )

( )

( )zx

zxzx

yzyz

yz

xyxy

xy

EG

EG

EG

τυτγ

τυτγ

τυτγ

+==

+==

+==

12

12

12

Generalized Hooke’s law

( )υ+=

12:_' EGlawsLAMÉ For isotropic materials

35


STEP 2: STRESS DETERMINATION AT A CERTAIN POINT The relation between the strain and the stress in the linear elastic domain is given by the generalized Hooke’s law:

( )( )

( )( )

( )

−

−

−

−−

−

+−=

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

E

γ

γ

γε

εε

υ

υ

υυυυ

υυυυυυ

υυ

τ

τ

τσ

σσ

22100000

02210000

00221000

000100010001

121

{ } [ ]{ }εσ D=

Date post:	19-Jan-2015
Category:	Technology
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Introduction to FEM

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